# Chemistry (Math in Disguise of Science)

**Flashcard maker :**Lily Taylor

2. Multiply the number they gave you over 1 GRAM and the number you got from adding the masses over 1 MOLE.

EX

Al-26.98

Cl-(3)35.45

133.33

1.25 mol/1 g X 133.33 g/1 mol= 166.62 g /mol

2. Multiply the number they gave you over 1 MOLECULE and the AVAGOTES number over the number you got.

EX

P-30.97

H-(3)1.0079

33.99 g

4.75g/1 Molecules X 6.22×10^23/33.99 g = 8.42×10^23 Molecules

2. Atomic mass g over the number you got.

3. Times 100

EX

C-12.01

H-(4)1.0079

16.04 g

C- 12.01gC/16.04g X 100= 75%

H- 4.03gH/16.04g X 100=25%

Baruim-89.59%

Oxygen-10.44%

Determine the empirical formula of it.

2. Put the number they gave you over 1 MOLE and multiply it by 1 MOLE over its atomic mass.

3. Put the number you got over the smallest number.

4. The number you now have is the subscript.

EX

Ba-89.59%=89.59g

O-10.44%=10.44g

89.59g Ba/ 1 mol X 1 mol/ 137.3 g = 0.6253 mol Ba

10.44g O/ 1 mol X 1 mol/16.00 g = 0.6253 mol O

0.6253 mol Ba/0.6253 mol =1 mol Ba

0.6253 mol O/0.6253 mol = 1 mol O

=

BaO

2. Divide both by the smallest number.

3. If you don’t get a full number , than multiply ALL till you do.

EX

Al- 4.151g Al/1 mol X 1 mol/26.98 g Al = .1538

O- 3.692g O/1 mol X 1 mol/ 16 G O= .23075

.1538/ .1538 = 1 X 2= 2

.23075/.1538=1.5 X 2=3

=Al2O3 ( Little 2 & Little 3)

2. Put the number they gave you over the number you got.

3. If you got number that info full multiply it till you do.

EX

C-12.01

H-(4)1.008

O-15.9994

32.0414

192g/32.0414g = 5.99= 6

C6H24O6 (little 4, little 24, little 6)

EX

6.22 g Al/ 1 mol X 1 mol/ 26.98 g Al = 0.23 g/mol Al

EX

Mg-24.305

N-(2)14.00674

O-(4)15.9994

116.31